\(A+H_2SO_4\rightarrow ASO_4+H_2\\ n_A=n_{H_2SO_4}=n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\b, M_A=\dfrac{5,6}{0,1}=56\left(\dfrac{g}{mol}\right)\Rightarrow A:Sắt\left(Fe=56\right)\\ a,PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ c,m_{H_2SO_4}=0,1.98=9,8\left(g\right)\\ C\%_{ddH_2SO_4}=\dfrac{9,8}{200}.100\%=4,9\%\)